Prove: if $Z_1$ and $Z_2$ are complex, then $Z_1*Z_2=R_1R_2[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)]$

given that $Z_n=R_n[\cos(\theta_N)+i\sin(\theta_N)]$

What I've written so far

proof: $Z_1*Z_2=R_1R_2*(cos\theta_1+isin\theta_1)(cos\theta_2+isin\theta_2)$

Which becomes the following:


And i was stuck at this point and do not know how to proceed. I would like a detailed explanation of how to proceed from here


Hint: What is $\cos \theta_1 + \theta_2$ and $\sin \theta_1+\theta_2$?

  • $\begingroup$ that comes from the generally accepted principle that if you multiply two complex numbers, then we would also add the angles as well which i am trying to prove. $\endgroup$ – John Rawls Feb 1 '17 at 2:55
  • $\begingroup$ @JohnRawls Actually, the trigonometric sum formulae can be easily derived from trigonometry only. You can search google for such a proof. $\endgroup$ – Teoc Feb 1 '17 at 2:57
  • $\begingroup$ i'm not sure if you would actually use trig identities in this proof verification problem, can you direct me how? $\endgroup$ – John Rawls Feb 1 '17 at 2:59
  • $\begingroup$ @JohnRawls You can use trigonometric identities to find that $\cos \theta_1+\theta_2=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2$ which is exactly the same as your real part. You can then proceed analogously for your imaginary part $\endgroup$ – Teoc Feb 1 '17 at 3:02

With the help of Display name I have proved it.

Recall the trig identity for addition formula to get the following:

$\cos \theta_1 + \theta_2$ and $\sin \theta_1+\theta_2$ becomes $cos\theta_1cos\theta_2-sin\theta_1sin\theta_2$ and $sin\theta_1cos\theta_2+sin\theta_2cos\theta_1$ respectively

so thus instead of starting from the left hand side of the proof, we shall start with the right hand side


which is = $R_1R_2(cos\theta_1+isin\theta_1)(cos\theta_2+isin\theta_2)$


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